Question

A metallic rod PQ whose length is $1 \, \text{m}$, is moving with a uniform speed of $2 \, \text{ms}^{-1}$ in a uniform magnetic field of $4 \, \text{T}$. A capacitor of $10 \, \mu \text{F}$ capacitance is connected as shown in the figure. Magnetic field is directed downwards, perpendicular to the plane of the paper. Find out:
Induced e.m.f. across the rod PQ.
Charge on the capacitor.
Which plate of the capacitor has positive charge?

Hint:

In motional emf: $E = B l v$. Always use Fleming’s right-hand rule to decide polarity.

Answer 1

Step 1: Formula for motional emf.
When a conductor of length $l$ moves with velocity $v$ perpendicular to magnetic field $B$: \[ E = B l v. \]
Step 2: Substitution.
\[ E = 4 \times 1 \times 2 = 8 \, \text{V}. \] So, the induced emf across PQ is $8 \, \text{V}$.
Step 3: Relation between charge and potential.
For capacitor: \[ Q = C V. \]
Step 4: Substitution.
\[ Q = (10 \times 10^{-6})(8) = 80 \times 10^{-6} \, \text{C}. \] \[ Q = 80 \, \mu \text{C}. \]
Step 5: Direction of charge flow.
Using Fleming’s right-hand rule: - Magnetic field ($B$) is into the page. - Velocity ($v$) is towards right. - Force on positive charge ($q$) is upwards (towards P). Thus, end $P$ becomes positive and end $Q$ becomes negative. Hence, plate $A$ connected to $P$ is positive.
Step 6: Conclusion.
(A) Induced emf $= 8 \, \text{V}$
(B) Charge on capacitor $= 80 \, \mu \text{C}$
(C) Plate $A$ is positive.